Question

$\int \frac{\tan 2\theta d\theta }{\sqrt{{\cos }^6\theta +{\sin }^6\theta }}$ is equal to

• A. $\ln \left|\frac{1+\sqrt{1+3{\cos }^{2}2\theta }}{\cos 2\theta }\right|+c$
• B. $\ln \left|\frac{\sqrt{1+3{\cos }^{2}2\theta }}{\cos 2\theta }\right|+c$
• C. $\ln \left|\frac{1+\sqrt{2+3{\cos }^{2}2\theta }}{\sin 2\theta }\right|+c$
• D. None of these

Answer 1

Answer: (A)

$\ln \left|\frac{1+\sqrt{1+3{\cos }^{2}2\theta }}{\cos 2\theta }\right|+c$

$I=\int \frac{\tan 2\theta d\theta }{\sqrt{{\cos }^6\theta +{\sin }^6\theta }}$

$=\int \frac{\sin 2\theta d\theta }{\cos 2\theta \sqrt{1-3{\sin }^2\theta {\cos }^2\theta }}$

$=\int \frac{\sin 2\theta d\theta }{\cos 2\theta \sqrt{1-\frac{3}{4}{\sin }^{2}2\theta }}$

$=\int \frac{2\sin 2\theta d\theta }{\cos 2\theta \sqrt{1+3{\cos }^{2}2\theta }}$
Put $\cos 2\theta =t$
$\Rightarrow -2\sin 2\theta d\theta =dt$

$\Rightarrow I=\int \frac{-dt}{t\sqrt{1+3t^2}}$
Put $1+3t^2=z^2$
$\Rightarrow 6tdt=2zdz$

So, $I=\int \frac{dz}{1-z^2}$

$=\frac{1}{2}\ln \left|\frac{1+z}{1-z}\right|+c_1$

$=\frac{1}{2}\ln \left|\frac{1+\sqrt{1+3t^2}}{1-\sqrt{1+3t^2}}\right|+c_1$

$=\ln \sqrt{\frac{1+\sqrt{1+3t^2}}{1-\sqrt{1+3t^2}}}+c_1$
$=\ln \left|\frac{1+\sqrt{1+3t^2}}{t}\right|+c$
$=\ln \left|\frac{1+\sqrt{1+3{\cos }^{2}2\theta }}{\cos 2\theta }\right|+c$